F(1) = (-4 sin(1), 2 cos(1) ) Use the closed path to find the "y'(t)" and "x(e)". x'(1) = -2 cos (1) y'(1) = -4 sin(1) x'(1) = -2 sin (1) y'(1) = -4 cos (1) B x'(1) = 2 cos (1) y'(1) = -4 sin (1) x'(1) = -2 sin (1) y'(1) = 4 cos (r) Now that you have found the parameterized vector field, "y't)", and "x'(t)" use them to find the dot product. F(1) - (v'(1), -x'(1) A F(1) - (v'(1), -x (1)) = 4 sin(1) cos (1) – 16 sin (1) cos (1) B F() - ('1), -x'(1)) = 8 cos°(1) + 32 sin'(1) F(1) · {v'(1), -x'(1) = - 8 cos(1) sin (0) O F(1) - (r'(1), -x'(1)) = -4 cos (1) + 8 sin (1) Now that we have found the value of our dot product, we can use it to find the flux of our situation. - / [F) - (v'1), -x'm>]dt and Imtervat: [0, 21] Flux = O / Fo. (v'(1), -r'1)>]dt = 0.56x
F(1) = (-4 sin(1), 2 cos(1) ) Use the closed path to find the "y'(t)" and "x(e)". x'(1) = -2 cos (1) y'(1) = -4 sin(1) x'(1) = -2 sin (1) y'(1) = -4 cos (1) B x'(1) = 2 cos (1) y'(1) = -4 sin (1) x'(1) = -2 sin (1) y'(1) = 4 cos (r) Now that you have found the parameterized vector field, "y't)", and "x'(t)" use them to find the dot product. F(1) - (v'(1), -x'(1) A F(1) - (v'(1), -x (1)) = 4 sin(1) cos (1) – 16 sin (1) cos (1) B F() - ('1), -x'(1)) = 8 cos°(1) + 32 sin'(1) F(1) · {v'(1), -x'(1) = - 8 cos(1) sin (0) O F(1) - (r'(1), -x'(1)) = -4 cos (1) + 8 sin (1) Now that we have found the value of our dot product, we can use it to find the flux of our situation. - / [F) - (v'1), -x'm>]dt and Imtervat: [0, 21] Flux = O / Fo. (v'(1), -r'1)>]dt = 0.56x
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
We are given a
![Vector Field: F(x,y) = (-x, y?) and
Closed Path: r(1) = (2 cos(r), -4 sin (r)>
Since we have our Path as a vector, we can use the General Formula to find Flux.
Use the closed path to convert the vector field into a parametric vector field.
(A) F(1) = (4 sin (r), 4 cos?(1))
B F(1) = (-2cos (1), 4 sin(r) >
© F(1) = (-2 cos (1), 16 sin²(1)>
D F(1) = (-4sản (1), 2 cos(r)>
Use the closed path to find the "y'(t)" and "x(t)".
x'(1) = -2 cos (1)
y'(r) = -4 sin (r)
x'(1) = -2 sin (1)
(B)
y'(r) = -4 cos ()
x'(1) = 2 cos ()
y'(r) = -4 sin (1)
D x'(1) = -2 sin (1)
y'(1) = 4 cos (r)
Now that you have found the parameterized vector field, "y'(t)", and "x'(t)" use them to find the dot product.
F(1) - (v'(1), -x'()>
(A
F(1) - (r'(1). -x'(1)) = 4 sin (t) cos (t) - 16 sin (t) cos (t)
(B
F(1) - ('(1). -x'(1)) = 8 cos (1) + 32 sin (2)
(c)
r'(1). -x'(1)) = - 8 cos (1) sin (t)
F(1)
D F(1) - ('(1). -x'(1)) = -4 cos²(1) + 8 sin²(1)
Now that we have found the value of our dot product, we can use it to find the flux of our situation.
F(1) - <v'(1), -x'(1)>]dt and intervat: [0, 2×]
Flux =
O Fo - (v'(). -x'()>]dt = 0.56x
Find the divergence of the vector field.
of
ду
A
v.F = -1
B
v.F = 5x + 2y2 - 8z
O v.F =0
O v.F = 4y - 5
Set up the triple integral for our path.
@ STI v.] de dy dz
Now that you have found the Divergence and set up your triple integral, we can use them with the Divergence Theorem to find the
Flux of our situation.
- SSS [v .F] av
@ SSS [v . F] av = 0
® SS[ [v F] av = 15
© J]/ [v F] a = - 125x
O SS[[v F] av = 75](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70427613-39b9-414c-b7e6-f1c6a3a1890c%2Fcecc896c-e64b-4a89-ae89-037466306fa3%2Fgfy0ayp_processed.png&w=3840&q=75)
Transcribed Image Text:Vector Field: F(x,y) = (-x, y?) and
Closed Path: r(1) = (2 cos(r), -4 sin (r)>
Since we have our Path as a vector, we can use the General Formula to find Flux.
Use the closed path to convert the vector field into a parametric vector field.
(A) F(1) = (4 sin (r), 4 cos?(1))
B F(1) = (-2cos (1), 4 sin(r) >
© F(1) = (-2 cos (1), 16 sin²(1)>
D F(1) = (-4sản (1), 2 cos(r)>
Use the closed path to find the "y'(t)" and "x(t)".
x'(1) = -2 cos (1)
y'(r) = -4 sin (r)
x'(1) = -2 sin (1)
(B)
y'(r) = -4 cos ()
x'(1) = 2 cos ()
y'(r) = -4 sin (1)
D x'(1) = -2 sin (1)
y'(1) = 4 cos (r)
Now that you have found the parameterized vector field, "y'(t)", and "x'(t)" use them to find the dot product.
F(1) - (v'(1), -x'()>
(A
F(1) - (r'(1). -x'(1)) = 4 sin (t) cos (t) - 16 sin (t) cos (t)
(B
F(1) - ('(1). -x'(1)) = 8 cos (1) + 32 sin (2)
(c)
r'(1). -x'(1)) = - 8 cos (1) sin (t)
F(1)
D F(1) - ('(1). -x'(1)) = -4 cos²(1) + 8 sin²(1)
Now that we have found the value of our dot product, we can use it to find the flux of our situation.
F(1) - <v'(1), -x'(1)>]dt and intervat: [0, 2×]
Flux =
O Fo - (v'(). -x'()>]dt = 0.56x
Find the divergence of the vector field.
of
ду
A
v.F = -1
B
v.F = 5x + 2y2 - 8z
O v.F =0
O v.F = 4y - 5
Set up the triple integral for our path.
@ STI v.] de dy dz
Now that you have found the Divergence and set up your triple integral, we can use them with the Divergence Theorem to find the
Flux of our situation.
- SSS [v .F] av
@ SSS [v . F] av = 0
® SS[ [v F] av = 15
© J]/ [v F] a = - 125x
O SS[[v F] av = 75
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